Problem source

\begin{questionparts}
\item Show, with the aid of  a sketch, that $y>  \tanh (y/2)$ for $y>0$ 
and deduce
that 
\begin{equation}
\textrm{arcosh} x > \dfrac{x-1}{\sqrt{x^2-1}} 
\text{  for  } x>1.
\tag{$*$}
\end{equation}
\item By integrating $(*)$, show that $\textrm{arcosh} x > 2
  \dfrac{{x-1}}{\sqrt{x^2-1}} $ for $x>1$.
\item Show that 
 $\textrm{arcosh} x >3
  \dfrac{\sqrt{x^2-1}}{{x+2}} 
$ for $x>1$.
\end{questionparts}
[\textbf{Note:} $\textrm{arcosh} x $ is another notation for
  $\cosh^{-1}x$.]

Solution source

\begin{questionparts}
\item \begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){tanh((#1)/2)};
    \def\xl{-2};
    \def\xu{2};
    \def\yl{-2};
    \def\yu{2};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=-2:2, samples=100] 
            plot (\x, {\functionf(\x)});

        \draw[thick, red] (-2,-2) -- (2,2);
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

If $y = \textrm{arcosh} x $, then $\tanh\textrm{arcosh} x/2 = \sqrt{\frac{\cosh \textrm{arcosh} x-1}{\cosh \textrm{arcosh} x+1}} = \sqrt{\frac{x-1}{x+1}} = \frac{x-1}{\sqrt{x^2-1}}$

\item \begin{align*}
\int \textrm{arcosh} x \d x &= \left [x \textrm{arcosh} x \right] - \int \frac{x}{\sqrt{x^2-1}} \d x \\
&= x \textrm{arcosh} x - \sqrt{x^2-1}+C \\
\int \frac{x-1}{\sqrt{x^2-1}} &= \sqrt{x^2-1} - \textrm{arcosh} x +C 
\end{align*}

Therefore

\begin{align*}
&& \int_1^x \textrm{arcosh} t \d t &> \int_1^x \frac{t-1}{\sqrt{t^2-1}} \d t \\
\Rightarrow && x \textrm{arcosh} x - \sqrt{x^2-1} - 0 &> \sqrt{x^2-1} - \textrm{arcosh} x  - 0 \\
\Rightarrow && (x+1) \textrm{arcosh} x &> 2\sqrt{x^2-1} \\
\Rightarrow && \textrm{arcosh} x & > 2\frac{\sqrt{x^2-1}}{x+1} \\
&&&= 2 \frac{\sqrt{x-1}}{\sqrt{x+1}} \\
&&&= 2 \frac{x-1}{\sqrt{x^2-1}}
\end{align*}

\item Integrating both sides again, 
\begin{align*}
&& \int_1^x  \textrm{arcosh} t \d t &> 2 \int_1^x \frac{t-1}{\sqrt{t^2-1}} \d t \\
\Rightarrow && x \textrm{arcosh} x - \sqrt{x^2-1}  &> 2\left (\sqrt{x^2-1} - \textrm{arcosh}x  \right) \\
\Rightarrow && (x+2)\textrm{arcosh} x &> 3 \sqrt{x^2-1} \\
\Rightarrow && \textrm{arcosh} x &> 3 \frac{\sqrt{x^2-1}}{x+2}
\end{align*}

\end{questionparts}